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The beta decay of cesium-137 has a half-life of 30.0 years. How many years must pass to reduce a 25.0 mg sample of cesium-137 to 9.38 mg?

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Answer:

t = 42.4 years

Step-by-step explanation:

To find the amount of time needed for the sample to decay, you need to use the half-life equation:


N(t) = N_0((1)/(2))^(t/h)

In this equation,

-----> N(t) = final mass (mg)

-----> N₀ = initial mass (mg)

-----> t = time passed (yrs)

-----> h = half-life (yrs)

You can find how much time passed by plugging the given variables into the equation and solving for "t". The final answer should have 3 sig figs like the given values.

N(t) = 9.38 mg t = ? yrs

N₀ = 25.0 mg h = 30.0 yrs


N(t) = N_0((1)/(2))^(t/h) <----- Half-life equation


9.38mg = 25.0mg((1)/(2))^(t/30.0yrs) <----- Insert variables


0.3752 = ((1)/(2))^(t/30.0yrs) <----- Divide both sides by 25.0 mg


ln(0.3752) = ln(((1)/(2))^(t/30.0yrs)) <----- Take the natural log of both sides


ln(0.3752) = (t)/(30.0yrs) ln((1)/(2)) <----- Rearrange the exponent


-9.803 = (t)/(30.0yrs) (-0.6931) <----- Solve the natural logs


1.1414= (t)/(30.0yrs) <----- Divide both sides by -0.6931


42.4 yrs= t <----- Multiply both sides by 30.0 yrs

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